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If an object leaves Earth with the escape velocity, meaning it'll never fall back on Earth, where does the energy go? You can't say it's converted to "potential energy" anymore because the object will never fall back. Yes, the escaping object velocity will decrease therefore energy will go somewhere else.

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The energy is indeed still converted to potential energy. The object doesn't go back on its own, but it still has a potential energy associated with its distance from the other mass.

Similarly, an object perfectly on the top of a hill won't roll down on its own, but it still has a potential energy associated with its being on top of the hill.

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But it is converted into potential energy. In non-general relativity terms, it is stored in the gravitational field surrounding and between the two objects (in general relativity terms, the energy is stored in how the curvature of space-time is altered).

In Newtonian gravity, you find the gravitational field of two point-like objects by adding the individual fields of the objects: $$\mathbf{g}_{\mathrm{tot}}(\mathbf{r}) = \mathbf{g}_{1}(\mathbf{r}) + \mathbf{g}_{2}(\mathbf{r}).$$

The energy density, $u_g(\mathbf{r})$ (energy per unit volume of space), stored in the gravitational field is given by $[G / 8\pi] \mathbf{g}_{\mathrm{tot}} \cdot \mathbf{g}_{\mathrm{tot}}$, giving: $$\begin{align} u_g(\mathbf{r}) &= \frac{G}{8\pi} [\mathbf{g}_{1}(\mathbf{r}) + \mathbf{g}_{2}(\mathbf{r})]\cdot [\mathbf{g}_{1}(\mathbf{r}) + \mathbf{g}_{2}(\mathbf{r})] \\ & = \frac{G}{8\pi}\left[\mathbf{g}_{1}^2 + \mathbf{g}_{2}^2 + 2 \mathbf{g}_{1} \cdot \mathbf{g}_{2} \right] \end{align}$$

The $\mathbf{g}_{1}^2$ and $\mathbf{g}_{2}^2$ terms are the density of energy needed to create objects 1 and 2, and add up to an energy that is infinite. The energy density of the interaction between 1 and 2 is given by the cross term: $$u_{g\, 1,2} = \frac{G}{4\pi} \mathbf{g}_{1} \cdot \mathbf{g}_{2}.$$ Since only differences in energy are relevant in classical Newtonian physics, the self energies don't contribute to the dynamics. If you integrate $u_{g\, 1,2}$ correctly over all space, you'll find that $$\Delta U = \frac{G m_1 m_2}{|\mathbf{r}_1 - \mathbf{r}_2|},$$ exactly as expected.

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There are three important things to note.
One is that you have simplified things by treating the Earth and point mass system as though there is nothing else in the Universe.
"Escape" means that of its own accord the point mass will never return back to the Earth but that does not mean that it ceases to exist.
Lastly you have implied that you can actually travel an infinite distance.

If the system is the mass under consideration and the Earth and there are no external force acting on the system (ie nothing else in the Universe) then as the separation between the mass and the Earth increases the kinetic energy of the system decreases at an every decreasing rate whilst at the same time the gravitational potential energy of the system increases at an ever decreasing rate.

If used in such a context, "infinity" is a word which means "the distance between the mass and the Earth is such that, on the scale set by the example, if there is a change in the distance between the mass and the Earth then the kinetic energy of the system can be sensibly taken to be zero and the gravitational potential energy can be sensibly taken to be constant.

So "mass at/reaches infinity" is a shorthand way of writing "the separation between the mass and the Earth tends to infinity" and this is a very useful idea when one starts to model a system mathematically as the limiting process is implied to have been performed - eg making one of the limits of an integration infinity.

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the energy after the object gone out of gravitational field of earth remain constant as it remains in equilibrium in space and does not fall to the earth afterward

  1. PE=mgh remain constant if h<=value of g at 'h' height)
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  • $\begingroup$ this isnt correct, your formula is an approximation for trajectories small enough for G to be constant. But G isnt constant . Plus my question mentionned escape velocity, meaning the object must moving away and not be "in equilibrium" $\endgroup$ – Manu de Hanoi May 3 at 20:56

protected by Qmechanic Dec 5 '16 at 23:26

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